Đáp án:$a.\left\{\begin{matrix}
x>0 & & \\
x\neq 1 & &
\end{matrix}\right.\\P=\frac{x-1}{\sqrt{x}}$
$b.P=\frac{3+2\sqrt{3}}{1+\sqrt{3}}$
$c.0<x<1$$
Giải thích các bước giải:
$a.DK:\left\{\begin{matrix}
x\geq 0 & & \\
\sqrt{x}-1\neq 0 & & \\
\sqrt{x}-x\neq 0 & & \\
\sqrt{x}+1\neq 0 & & \\
x-1\neq 0 & &
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x>0 & & \\
x\neq 1 & &
\end{matrix}\right.\\
P=\left (\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-x} \right ):\left ( \frac{1}{\sqrt{x}+1}+\frac{2}{x-1} \right )\\
=\left ( \frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}.(\sqrt{x}-1)}-\frac{1}{\sqrt{x}.(\sqrt{x}-1)} \right ):\left ( \frac{\sqrt{x}-1}{(\sqrt{x}+1).(\sqrt{x}-1)}+\frac{2}{(\sqrt{x}+1)(\sqrt{x}-1)} \right )\\
=\frac{x-1}{\sqrt{x}.(\sqrt{x}-1)}:\frac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}\\
=\frac{x-1}{\sqrt{x}}$
$b.x=2(\sqrt{3}+2)=4+2\sqrt{3}\\
\Rightarrow P=\frac{4+2\sqrt{3}-1}{\sqrt{4+2\sqrt{3}}}=\frac{3+2\sqrt{3}}{\sqrt{(1+\sqrt{3})^2}}=\frac{3+2\sqrt{3}}{1+\sqrt{3}}$
$c.P=\frac{x-1}{\sqrt{x}}<0\\
\sqrt{x}> 0\Rightarrow P<0\Leftrightarrow x-1<0\Rightarrow x<1$
Đối chiếu điều kiện ⇒$0<x<1$
